Answer to Question #213132 in Statistics and Probability for Dr. Nazir

Question

The finished inside diameter of a piston ring is normally distributed with a mean of 10 centimeters and a standard deviation of 0.03 centimeter.

(a) What proportion of rings will have inside diameters exceeding 10.075 centimeters?

(b) What is the probability that a piston ring will have an inside diameter between 9.97 and 10:03 centimeters?

(c) Below what value of inside diameter will 15% of the piston rings fall?

Answer 1

Let X represent that the finished inside diameter of a piston ring is normally distributed with the mean 10 centimeters and a standard deviation of 0.03 centimeters.

(a) The proportion of rings will have inside diameters exceeding 10.075 centimeters

$z=\frac{x- \mu}{\sigma}= \frac{10.075-10}{0.03}=2.5 \\ P(X>10.075) = P(Z>2.5) \\ = 1 - P(Z<2.5) \\ = 1 -0.9938 \\ = 0.0062 = 0.62 \; \%$

(b) The probability that a piston ring will have an inside diameter between 9.97 and 10:03 centimeters

$P(9.97<X<10.03) = P(X<10.03) -P(X<9.97) \\ =P(Z< \frac{10.03-10}{0.03}) -P(Z< \frac{9.97-10}{0.03}) \\ =P(Z< 1) -P(Z< -1) \\ = 0.8413 -0.1587 \\ = 0.6826$

(c) The value of inside diameter that will 15% of the piston rings fall

$P(X<x) = 0.15 \\ P(Z< \frac{x-10}{0.03}) = 0.15 \\ P(Z< -1.036) = 0.15 \\ \frac{x-10}{0.03}= -1.036 \\ x-10 = - 1.036 \times 0.03 \\ x= -1.036 \times 0.03 + 10 \\ x = 9.9689$